Solving De Broglie Wave: Direction & F/G

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The discussion revolves around the de Broglie wave function, expressed as ψ = e^{i(px-Et)/ħ, with participants analyzing its direction and calculating derivatives F and G. It is concluded that the wave travels in the positive direction as time increases. Participants evaluate the derivatives, confirming that F(x,t) and G(x,t) are correctly derived, though initial expressions were confusing. The calculation of F/G resulting in 2m/ħ is questioned for accuracy, with a focus on its significance in the context of wave equations in potential-free regions. The conversation emphasizes the importance of clarity in notation and understanding partial derivatives in wave mechanics.
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the de Broglie wave has the form: /psi = e^{i(px-Et)/ \hbar}

1. I am asked the direction of this wave. To me, as t increases, the x has to increase as well to keep a constant function on the left hand side so I believe that the direction of this wavelangth is heading towards the positive end.
2. I am to find:

F(x,t)=-\frac{\delta ^2 \phi}{\delta x^2}
and
G(x,t)=i\frac{\delta \phi}{\delta t}

i'm not too good with partial derivatives since I just learned them a week ago, please correct me if I am wrong.

x and t don't depend on each other while E and p are fixed energy and momentum:

first partial derivative:
F=\frac{ip}{ \hbar} e^{i(px-Et)/ \hbar}
second:
F=-\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}
F(x,t)=\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}

first partial derivative:
G=-\frac{iE}{\hbar} e^{i(px-Et)/ \hbar}
G(x,t)=\frac{E}{\hbar} e^{i(px-Et)/ \hbar}

have I done this correctly?

then I am to evaluate F/G knowing that E=p^2/2m

I get \frac{F}{G}=\frac{2m}{\hbar}?
 
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UrbanXrisis said:
first partial derivative:
F=\frac{ip}{ \hbar} e^{i(px-Et)/ \hbar}
second:
F=-\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}
F(x,t)=\frac{p^2}{ \hbar ^2} e^{i(px-Et)/ \hbar}
Here you have three different expressions for F. So which on is it?

I understand what you're doing, but as you've written it down is confusing (and wrong). Your final result for F(x,t) is correct though, which was what you're heading for.
first partial derivative:
G=-\frac{iE}{\hbar} e^{i(px-Et)/ \hbar}
G(x,t)=\frac{E}{\hbar} e^{i(px-Et)/ \hbar}
Same thing here. The first expression is not G, but \frac{\partial \phi}{\partial t}, so you should write that.

The rest looks good.

then I am to evaluate F/G knowing that E=p^2/2m

I get \frac{F}{G}=\frac{2m}{\hbar}?[/QUOTE]
 
In the LaTeX, use \ before psi, and use \del or \partial for the partial derivative.

For partial differentiation, one treats the other variable as a contant, and what you have done appears to be correct, except as Galileo pointed out, write the partials of F and G where you are differentiating.

The wave is in the positive direction.

Another way to write the wave equation is \psi(x,t)\,=\,A\,exp\,[i(kx-\omega t)], where k\,=\,2\pi/\lambda and \omega\,=\,2\pi\nu.
 
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for F and G, I was just showing the first partial derivative and the second partial derivative.

was my calculation of \frac{F}{G}=\frac{2m}{\hbar} correct?

what is the significance of F/G?
 
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In one dimension -

\frac{d^2\psi}{dx^2}\,+\,\frac{2mE}{\hbar^2}\psi\,=\,0

in region with the potential, V(x) = 0.
 
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